Mathematics 9709/33 May/June 2020: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 3 Pure Mathematics 3, May/June 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
1 State or imply non-modular inequality (2x – 1)2 > 32(x + 2)2, or corresponding quadratic equation, or pair of linear equations B1
Make reasonable attempt at solving a 3-term quadratic, or solve two linear equations for x M1
Obtain critical values x = –7 and x = –1 A1
State final answer –7 < x < –1 A1
Alternative method for question 1
Obtain critical value x = –1 from a graphical method, or by solving a linear equation or linear inequality B1
Obtain critical value x = –7 similarly B2
State final answer –7 < x < –1 B1
[Do not condone ⩽ for < in the final answer.]
4
2 Commence integration and reach M1*
a (2 − x)e
−2 x −2 x
+b e dx , or equivalent
1 1 A1
Obtain −
2
( 2 − x ) e−2 x − e−2 x dx , or equivalent
2
1 1 A1
Complete integration and obtain − ( 2 − x ) e−2 x + e−2 x , or equivalent
2 4
Use limits correctly, having integrated twice DM1
1 A1
Obtain answer
4
( )
3 − e−2 , or exact equivalent
5
(a) Show that the equation ln 1 + e−x + 2x = 0 can be expressed as a quadratic equation in ex . [2]
(b) Hence solve the equation ln 1 + e−x + 2x = 0, giving your answer correct to 3 decimal places.
[4]
Mark scheme
3(a) Remove logarithms correctly and state 1 + e –x = e –2x, or equivalent B1
Show equation is + − 1 = 0, where = e , or equivalent B1
2
3(b) Solve a 3-term quadratic for u M1
1 A1
Obtain root
2
( )
−1 + 5 , or decimal in [0.61, 0.62]
Use correct method for finding x from a positive root M1
Obtain answer x = –0.481 only A1
4
The equation of a curve is y = x tan−1 12 x .
dy
(a) Find . [3]
dx
(b) The tangent to the curve at the point where x = 2 meets the y-axis at the point with coordinates
0, p.
Find p. [3]
Mark scheme
4(a) Use the product rule M1
State or imply derivative of tan ( ) is of the form /(4 + ), where k = 2 or 4, or equivalent M1
1 2x A1
Obtain correct derivative in any form, e.g. tan −1 x + 2 , or equivalent
2 x +4
3
4(b) 1 B1
State or imply y-coordinate is π
2
Carry out a complete method for finding p, e.g. by obtaining the equation of the tangent and setting x = 0, or by equating the M1
1
π−p
gradient at x = 2 to 2
2
Obtain answer p = –1 A1
3
By first expressing the equation tan 1 tan 1 + 45Å = 2 cot 21 as a quadratic equation in tan 1, solve the equation for 0Å < 1 < 90Å. [6]
Mark scheme
5 Use tan 2A formula to express RHS in terms of tan θ M1
Use tan (A ± B) formula to express LHS in terms of tan θ M1
Using tan 45° = 1, obtain a correct horizontal equation in any form A1
Reduce equation to 2 tan 2 θ + tan θ − 1 = 0 A1
Solve a 3-term quadratic and find a value of θ M1
Obtain answer θ = 26.6° and no other A1
6
(a) By sketching a suitable pair of graphs, show that the equation x5 = 2 + x has exactly one real root. [2]
(b) Show that if a sequence of values given by the iterative formula
4xn5 + 2 xn+1 =
5xn4 − 1
converges, then it converges to the root of the equation in part (a). [2]
(c) Use the iterative formula with initial value x1 = 1.5 to calculate the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]
2
Mark scheme
6(a) Sketch a relevant graph, e.g. y = x5 B1
Sketch a second relevant graph, e.g. y = x + 2 and justify the given statement B1
2
6(b) 4 x5 + 2 B1
State a suitable equation, e.g. x =
5x4 − 1
Rearrange this as x5 = 2 + x or commence working vice versa B1
2
6(c) Use the iterative formula correctly at least once M1
Obtain final answer 1.267 A1
Show sufficient iterations to 5 d.p. to justify 1.267 to 3 d.p., or show there is a sign change in the interval (1.2665, 1.2675) A1
3
Let f x = .
2x − 1 2x + 1
(a) Express f x in partial fractions. [2]
(b) Using your answer to part (a), show that
2 1 1 1 1 f x = − + + . [2]
2x − 12 2x − 1 2x + 1 2x + 12
2 2
(c) Hence show that Ó f x dx = 25 + 12 ln 59 . [5]
1
Mark scheme
7(a) A B M1
State or imply the form + and use a relevant method to find A or B
2x − 1 2x + 1
Obtain A = 1, B = –1 A1
2
7(b) Square the result of part (a) and substitute the fractions of part (a) M1
Obtain the given answer correctly A1
2
7(c) 1 1 1 1 B3, 2, 1, 0
Integrate and obtain − − ln ( 2 x − 1) + ln ( 2 x + 1) − , or equivalent
2 ( 2 x − 1) 2 2 2 ( 2 x + 1)
Substitute limits correctly M1
Obtain the given answer correctly A1
5
ρ ρ
Relative to the origin O, the points A, B and D have position vectors given by
OA = i + 2j + k, OB = 2i + 5j + 3k and OD = 3i + 2k.
A fourth point C is such that ABCD is a parallelogram.
(a) Find the position vector of C and verify that the parallelogram is not a rhombus. [5]
(b) Find angle BAD, giving your answer in degrees. [3]
(c) Find the area of the parallelogram correct to 3 significant figures. [2]
Mark scheme
8(a) State or imply AB or AD in component form B1
Use a correct method for finding the position vector of C M1
Obtain answer 4i + 3j + 4k, or equivalent A1
Using the correct process for the moduli, compare lengths of a pair of adjacent sides, e.g. AB and AD M1
Show that ABCD has a pair of unequal adjacent sides A1
Alternative method for question 8(a)
ρ ρ
State or imply AB or AD in component form B1
Use a correct method for finding the position vector of C M1
Obtain answer 4i + 3j + 4k, or equivalent A1
ρ ρ
Use the correct process to calculate the scalar product of AC and BD , or equivalent M1
Show that the diagonals of ABCD are not perpendicular A1
5
ρ ρ
8(b) Use the correct process to calculate the scalar product of a pair of relevant vectors, e.g. AB and AD M1
Using the correct process for the moduli, divide the scalar product by the product of the moduli of the two vectors and evaluate M1
the inverse cosine of the result
Obtain answer 100.3° A1
3
8(c) Use a correct method to calculate the area, e.g. calculate AB.AC sin BAD M1
Obtain answer 11.0 A1 FT
(FT on angle BAD)
2
(a) The complex numbers u and w are such that u − w = 2i and uw = 6.
Find u and w, giving your answers in the form x + iy, where x and y are real and exact. [5]
(b) On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying the inequalities
z − 2 − 2i ≤ 2, 0 ≤ arg z ≤ 14 π and Re z ≤ 3. [5]
Mark scheme
9(a) Eliminate u or w and obtain an equation w or u M1
Obtain a quadratic in u or w, e.g. u2 – 2iu – 6 = 0 or w2 + 2iw – 6 = 0 A1
Solve a 3-term quadratic for u or for w M1
Obtain answer u = 5 + i , w = 5 − i A1
Obtain answer u = − 5 + i , w = − 5 − i A1
5
9(b) Show the point representing 2 + 2i B1
Show a circle with centre 2 + 2i and radius 2 B1 FT
(FT is on the position of 2 + 2i)
Show half-line from origin at 45° to the positive x-axis B1
Show line for Re z = 3 B1
Shade the correct region B1
5
r
h
A
A tank containing water is in the form of a hemisphere. The axis is vertical, the lowest point is A and the radius is r , as shown in the diagram. The depth of water at time t is h. At time t = 0 the tank is water is r. At this instant a tap at A is opened and water begins to flow out at full and the depth of the a rate proportional to h. The tank becomes empty at time t = 14.
The volume of water in the tank is V when the depth is h. It is given that V = 13 π 3rh2 − h3 .
(a) Show that h and t satisfy a differential equation of the form dh B
=− ,
2rh 2 − h 2
1 3 dt
where B is a positive constant. [4]
(b) Solve the differential equation and obtain an expression for t in terms of h and r. [8]
Additional Page
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
10(a) dV B1
State or imply = −k h
dt
dV B1
State or imply = 2πrh − πh 2 , or equivalent
dh
dV dV dh M1
Use = .
dt dh dt
Obtain the given answer correctly A1
4
10(b) Separate variables and attempt integration of at least one side M1
3
4 2 2 2
5 A3, 2, 1, 0
Obtain terms rh − h and –Bt
3 5
Use t = 0, h = r to find a constant of integration c M1
Use t = 14, h = 0 to find B M1
5
14 2 1
5 A1
Obtain correct c and B, e.g. c = r , B = r2
15 15
3 5 A1
h 2 h 2
Obtain final answer t = 14 − 20 + 6 , or equivalent
r r
8
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/33, May/June 2020 (9709_s20_ms_33), and the question paper (9709_s20_qp_33), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/33 for May/June 2020 is out of 75 marks, with 1 hour 50 minutes to answer it.
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